# hw5 - johnson(rj6247 – hw 5 – Opyrchal –(121014 1...

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Unformatted text preview: johnson (rj6247) – hw 5 – Opyrchal – (121014) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 75 m length of coaxial cable has a solid cylindrical wire inner conductor with a di- ameter of 3 . 142 mm and carries a charge of 7 . 08 μ C. The surrounding conductor is a cylindrical shell and has an inner diameter of 7 . 928 mm and a charge of- 7 . 08 μ C. Assume the region between the conductors is air. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . What is the capacitance of this cable? Correct answer: 4 . 50811 nF. Explanation: Let : k e = 8 . 98755 × 10 9 N · m 2 / C 2 , Q = 7 . 08 μ C , ℓ = 75 m , a = 3 . 142 mm , and b = 7 . 928 mm . The charge per unit length is λ ≡ Q ℓ . V =- integraldisplay b a vector E · dvectors =- 2 k e λ integraldisplay b a dr r =- 2 k e Q ℓ ln parenleftbigg b a parenrightbigg . The capacitance of a cylindrical capacitor is given by C ≡ Q V = ℓ 2 k e 1 ln parenleftbigg b a parenrightbigg = 75 m 2 (8 . 98755 × 10 9 N · m 2 / C 2 ) × 1 ln parenleftbigg 7 . 928 mm 3 . 142 mm parenrightbigg · parenleftbigg 1 × 10 9 nF 1 F parenrightbigg = 4 . 50811 nF 002 (part 2 of 2) 10.0 points What is the potential difference between the two conductors? Correct answer: 1 . 5705 kV. Explanation: Therefore V = Q C = 7 . 08 μ C 4 . 50811 nF parenleftbigg 1 × 10 9 nF 1 F parenrightbigg × parenleftbigg 1 C 1 × 10 6 μ C parenrightbiggparenleftbigg 1 kV 1000 V parenrightbigg = 1 . 5705 kV 003 10.0 points Two conductors insulated from each other are charged by transferring electrons from one conductor to the other. After 2 . 621 × 10 13 have been transferred, the potential difference between the conductors is 14 . 6 V. The charge on an electron is- 1 . 60218 × 10 − 19 C . What is the capacitance of the system? Correct answer: 2 . 87624 × 10 − 7 F. Explanation: Let : e =- 1 . 60218 × 10 − 19 C , N = 2 . 621 × 10 13 electrons , and V = 14 . 6 V . One conductor gains a charge of Q =- N e while the other gains a charge of- Q. The capacitance of the system is C = q V =- N e V =- (2 . 621 × 10 13 ) (- 1 . 60218 × 10 − 19 C) 14 . 6 V = 2 . 87624 × 10 − 7 F . johnson (rj6247) – hw 5 – Opyrchal – (121014) 2 004 (part 1 of 2) 10.0 points The potential difference between a pair of oppositely charged parallel plates is 386 V....
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## This note was uploaded on 04/03/2009 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.

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hw5 - johnson(rj6247 – hw 5 – Opyrchal –(121014 1...

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